The point was: Why use MAPLE?? You can't PROVE the MAPLE calculation is correct by using MAPLE!
It's easy to calculate that the derivative of
tan3(x)- 3tan(x) is 3tan(x)sec2(x)- 3sec2(x) and certainly you should know that [itex]tan(\frac{\pi}{3})= \sqrt{3}[/itex] and that [itex]sec(\frac{\pi}{3}})= 2[/itex]. Of course, the derivative of [itex]cos(x+\frac{\pi}{6}[/itex] is [itex]-sin(x+\frac{\pi}{6})[/itex] and [itex]-sin(\frac{\pi}{2}})= -1[/itex].

The point was: Why use MAPLE?? You can't PROVE the MAPLE calculation is correct by using MAPLE!
It's easy to calculate that the derivative of
tan3(x)- 3tan(x) is 3tan(x)sec2(x)- 3sec2(x) and certainly you should know that [itex]tan(\frac{\pi}{3})= \sqrt{3}[/itex] and that [itex]sec(\frac{\pi}{3}})= 2[/itex]. Of course, the derivative of [itex]cos(x+\frac{\pi}{6}[/itex] is [itex]-sin(x+\frac{\pi}{6})[/itex] and [itex]-sin(\frac{\pi}{2}})= -1[/itex].

The limit, by L'Hopital, is (3*3*4- 3(22)/-1= -(36-12)= -24.

Click to expand...

That's indeed correct (notice that there's a typo, it should obviously be

"It's easy to calculate that the derivative of
tan3(x)- 3tan(x) is 3tan2(x)sec2(x)- 3sec2(x)"

i.e. there is a square in the first tan in the derivative but it's obviously a typo because HallsofIvy gave the correct numerical answer.

Just a thought: if someone forgets to square the sec(x), they will get -12. So that could be the origin of the incorrect answer -12.